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Leokris [45]
3 years ago
9

A solid metal sphere of radius a = 2.5 cm has a net charge Qin = - 3 nC (1 nC = 10-9C). The sphere is surrounded by a concentric

conducting spherical shell of inner radius b = 6 cm and outer radius c = 9 cm. The shell has a net charge Qout = + 2 nC. What is V0, the electric potential at the center of the metal sphere, given the potential at infinity is zero? V0 =
Physics
1 answer:
Furkat [3]3 years ago
4 0

Answer: 630 V

Explanation: In order to solve this problem we have to consider the potential given by:

In the region 0<a<b

V(r)= -∫E*dr where E can be calculated by Gaussian law then E= k*qa/r^2

where qa=-3 nC

then

The V(r)=k*qa/r+C C is zero since V(r=∞)=0

Finally we have

V(r)= k*qa (1/r)-(1/b)

thus V(a)= k*qa (1/a)-(1/b)

Replacing the values V(a) = 630 V, as the solid metal sphere is a equipotential object thus at the center have the same value of V that in r=a ( 630 V).

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One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = \sqrt{6gL}

Hence, v0 = 6.86 m/s

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